JEE Main 2022 June 27 – Shift 1 Maths Question Paper with Solutions

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SECTION – A

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

Choose the correct answer :

1. The area of the polygon, whose vertices are the non-real roots of the equation

\begin{array}{l} \overset{―}{z} = i z^{2} \end{array}

is :

\begin{array}{l} (A) \frac{3 \sqrt{3}}{4} \end{array}

\begin{array}{l} (B) \frac{3 \sqrt{3}}{2} \end{array}

\begin{array}{l} (C) \frac{3}{2} \end{array}

\begin{array}{l} (D) \frac{3}{4} \end{array}

Answer (A)

Sol.

\begin{array}{l} \overset{―}{z} = i z^{2} \end{array}

Let z = x + iy

x – iy = i(x² – y² + 2xiy)

x – iy = i(x² – y²) – 2xy

∴ x = –2yx or x² – y² = –y

x = 0 or

\begin{array}{l} y = - \frac{1}{2} \end{array}

Case-I

x = 0

–y² = –y

y = 0, 1

Case-II

\begin{array}{l} y = - \frac{1}{2} \end{array}

⇒

\begin{array}{l} x^{2} - \frac{1}{4} = \frac{1}{2} \Rightarrow x = \pm \frac{\sqrt{3}}{2} \end{array}

\begin{array}{l} z = {0, i, \frac{\sqrt{3}}{2} - \frac{i}{2}, \frac{- \sqrt{3}}{2} - \frac{i}{2}} \end{array}

Area of polygon

\begin{array}{l} = \frac{1}{2} | \begin{array}{c} 0 & 1 & 1 \\ \frac{\sqrt{3}}{2} & \frac{- 1}{2} & 1 \\ \frac{- \sqrt{3}}{2} & \frac{- 1}{2} & 1 \end{array} | \end{array}

\begin{array}{l} = \frac{1}{2} | - \sqrt{3} - \frac{\sqrt{3}}{2} | = \frac{3 \sqrt{3}}{4} \end{array}

2. Let the system of linear equations x + 2y + z = 2, αx + 3y – z = α, –αx + y + 2z = –α be inconsistent. Then α is equal to :

\begin{array}{l} (A) \frac{5}{2} \end{array}

\begin{array}{l} (B) - \frac{5}{2} \end{array}

\begin{array}{l} (C) \frac{7}{2} \end{array}

\begin{array}{l} (D) - \frac{7}{2} \end{array}

Answer (D)

Sol. x + 2y + z = 2

αx + 3y – z = α

–αx + y + 2z = –α

\begin{array}{l} Δ = | \begin{array}{c} 1 & 2 & 1 \\ α & 3 & - 1 \\ - α & 1 & 2 \end{array} | = 1 (6 + 1) - 2 (2 α - α) + 1 (α + 3 α) \end{array}

= 7 + 2α

\begin{array}{l} Δ = 0 \Rightarrow α = - \frac{7}{2} \end{array}

\begin{array}{l} Δ_{1} = | \begin{array}{c} 2 & 2 & 1 \\ α & 3 & - 1 \\ - α & 1 & 2 \end{array} | = 14 + 2 α \neq 0 for α = - \frac{7}{2} \end{array}

∴ For no solution

\begin{array}{l} α = - \frac{7}{2} \end{array}

3. If

\begin{array}{l} x = \sum_{n = 0}^{\infty} a^{n}, y = \sum_{n = 0}^{\infty} b^{n}, z = \sum_{n = 0}^{\infty} c^{n}, \end{array}

where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc≠ 0,

then :

(A) x, y, zare in A.P.

(B) x, y, zare in G.P.

(C)

\begin{array}{l} \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \end{array}

are in A.P.

(D)

\begin{array}{l} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 - (a + b + c) \end{array}

Answer (C)

Sol.

\begin{array}{l} x = \sum_{n = 0}^{\infty} a^{n} = \frac{1}{1 - a}; y = \sum_{n = 0}^{\infty} b^{n} = \frac{1}{1 - b}; z = \sum_{n = 0}^{\infty} c^{n} = \frac{1}{1 - c} \end{array}

Now,

a, b, c→ AP

1 – a, 1 – b, 1 – c→ AP

\begin{array}{l} \frac{1}{1 - a}, \frac{1}{1 - b}, \frac{1}{1 - c} \to HP \end{array}

x, y, z→ HP

∴ [tex]\frac{1}{x},\frac{1}{y},\frac{1}{z}\rightarrow\textup{AP}[/tex]

4. Let

\begin{array}{l} \frac{d y}{d x} = \frac{a x - b y + a}{b x + c y + a} \end{array}

where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is

(A) 10

(B) 8

(C) 7

(D) 5

Answer (B)

Sol.

\begin{array}{l} \frac{d y}{d x} = \frac{a x - b y + a}{b x + c y + a} \end{array}

= bxdy + cy dy + a dy = ax dx – by dx + a dx

= cy dy + a dy – ax dx – a dx + b(x dy + y dx) = 0

\begin{array}{l} = c \int y d y + a \int x d x - a \int d x + b \int d (x y) = 0 \end{array}

\begin{array}{l} = \frac{c y^{2}}{2} + a y - \frac{a x^{2}}{2} - a x + b x y = k \end{array}

= ax² – cy² + 2ax – 2ay – 2bxy = k

Above equation is circle

⇒ a = –c and b = 0

ax² + ay² + 2ax – 2ay = k

⇒ x² + y² + 2x – 2y = λ

\begin{array}{l} [λ = \frac{k}{a}] \end{array}

Passes through (2, 5)

4 + 25 + 4 – 10 = λ⇒λ = 23

Circle ≡x² + y² + 2x – 2y – 23 = 0

Centre (–1, 1)

\begin{array}{l} r = \sqrt{{(- 1)}^{2} + 1^{2} + 23 = 5} \end{array}

Shortest distance of (11, 6)

\begin{array}{l} = \sqrt{12^{2} + 5^{2}} - 5 \end{array}

= 13 – 5

= 8

5. Let a be an integer such that

\begin{array}{l} lim_{x \to 7} \frac{18 - [1 - x]}{[x - 3 a]} \end{array}

exists, where [t] is greatest integer ≤ t. Then a is equal

to :

(A) –6

(B) –2

(C) 2

(D) 6

Answer (A)

Sol.

\begin{array}{l} lim_{x \to 7} \frac{18 - [1 - x]}{[x - 3 a]} \end{array}

exist &a∈I.

\begin{array}{l} = lim_{x \to 7} \frac{17 - [- x]}{[x] - 3 a} \end{array}

exist

\begin{array}{l} R H L = lim_{x \to 7^{+}} \frac{17 - [- x]}{[x] - 3 a} = \frac{25}{7 - 3 a} [a \neq \frac{7}{3}] \end{array}

\begin{array}{l} L H L = lim_{x \to 7^{-}} \frac{17 - [- x]}{[x] - 3 a} = \frac{24}{6 - 3 a} [a \neq 2] \end{array}

For limit to exist

LHL = RHL

\begin{array}{l} \frac{25}{7 - 3 a} = \frac{24}{6 - 3 a} \end{array}

\begin{array}{l} \Rightarrow \frac{25}{7 - 3 a} = \frac{8}{2 - a} \end{array}

∴ _{a = -6}

6. The number of distinct real roots of x⁴ – 4x + 1 = 0 is :

(A) 4

(B) 2

(C) 1

(D) 0

Answer (B)

Sol.

ƒ(x) = x⁴ – 4x + 1 = 0

ƒ′(x) = 4x³ – 4

= 4(x – 1) (x² + 1 + x)

JEE Main 2022 June 27 Shift 1 Maths A6

⇒ Two solution

7. The lengths of the sides of a triangle are 10 + x², 10 + x² and 20 – 2x². If for x = k, the area of the triangle is maximum, then 3k² is equal to :

(A) 5

(B) 8

(C) 10

(D) 12

Answer (C)

Sol.

\begin{array}{l} C D = \sqrt{{(10 + x^{2})}^{2} - {(10 - x^{2})}^{2}} = 2 \sqrt{10} | x | \end{array}

Area

\begin{array}{l} = \frac{1}{2} \times C D \times A B = \frac{1}{2} \times 2 \sqrt{10} | x | (20 - 2 x^{2}) \end{array}

\begin{array}{l} A = \sqrt{10} | x | (10 - x^{2}) \end{array}

\begin{array}{l} \frac{d A}{d x} = \sqrt{10} \frac{| x |}{x} (10 - x^{2}) + \sqrt{10} | x | (- 2 x) = 0 \end{array}

⇒ 10 – x² = 2x²

3x² = 10

x = k

3k² = 10

8. If

\begin{array}{l} \cos^{- 1} (\frac{y}{2}) = {log}_{e} {(\frac{x}{5})}^{5}, | y | < 2 \end{array}

then :

(A) x²y′′ + xy′ – 25y = 0

(B) x²y′′ – xy′ – 25y = 0

(C) x²y′′ – xy′+ 25y = 0

(D) x²y′′ + xy′+ 25y = 0

Answer (D)

Sol.

\begin{array}{l} \cos^{- 1} (\frac{y}{2}) = {log}_{e} {(\frac{x}{5})}^{5}, | y | < 2 \end{array}

Differentatingon both side

\begin{array}{l} - \frac{1}{\sqrt{1 - {(\frac{y}{2})}^{2}}} \times \frac{y^{‘}}{2} = \frac{5}{\frac{x}{5}} \times \frac{1}{5} \end{array}

\begin{array}{l} \frac{- x y^{‘}}{2} = 5 \sqrt{1 - {(\frac{y}{2})}^{2}} \end{array}

Square on both side

\begin{array}{l} \frac{x^{2} y^{‘ 2}}{4} = 25 (\frac{4 - y^{2}}{4}) \end{array}

Diff on both side

\begin{array}{l} 2 x y^{‘ 2} + 2 y^{‘} y^{”} x^{2} = - 25 \times 2 y y^{‘} \end{array}

xy′ + y′′x² + 25y = 0

9. If

\begin{array}{l} \int \frac{(x^{2} + 1) e^{x}}{{(x + 1)}^{2}} d x = f (x) e^{x} + C \end{array}

where C is a constant, then

\begin{array}{l} \frac{d^{3} f}{d x^{3}} \end{array}

at x = 1 is equal to :

\begin{array}{l} (A) - \frac{3}{4} \end{array}

\begin{array}{l} (B) \frac{3}{4} \end{array}

\begin{array}{l} (C) - \frac{3}{2} \end{array}

\begin{array}{l} (D) \frac{3}{2} \end{array}

Answer (B)

Sol.

\begin{array}{l} I = \int \frac{e^{x} (x^{2} + 1)}{{(x + 1)}^{2}} d x = f (x) e^{x} + c \end{array}

\begin{array}{l} I = \int \frac{e^{x} (x^{2} - 1 + 1 + 1)}{{(x + 1)}^{2}} d x \end{array}

\begin{array}{l} = \int e^{x} [\frac{x - 1}{x + 1} + \frac{2}{{(x + 1)}^{2}}] d x \end{array}

\begin{array}{l} = e^{x} (\frac{x - 1}{x + 1}) + c \end{array}

\begin{array}{l} ∴ f (x) = \frac{x - 1}{x + 1} \end{array}

\begin{array}{l} f (x) = 1 - \frac{2}{x + 1} \end{array}

\begin{array}{l} f^{'} (x) = 2 {(\frac{1}{x + 1})}^{2} \end{array}

\begin{array}{l} f ” (x) = - 4 {(\frac{1}{x + 1})}^{3} \end{array}

\begin{array}{l} f ”^{'} (x) = \frac{12}{{(x + 1)}^{4}} \end{array}

for x = 1

\begin{array}{l} f ”^{'} (1) = \frac{12}{2^{4}} = \frac{12}{16} = \frac{3}{4} \end{array}

10. The value of the integral

\begin{array}{l} \int_{- 2}^{2} \frac{| x^{3} + x |}{(e^{x | x |} + 1)} d x \end{array}

is equal to:

(A) 5e²

(B) 3e^–2

(C) 4

(D) 6

Answer (D)

Sol.

\begin{array}{l} I = \int_{- 2}^{2} \frac{| x^{3} + x |}{e^{x | x |} + 1} d x \dots . (i) \end{array}

\begin{array}{l} I = \int_{- 2}^{2} \frac{| x^{3} + x |}{e^{- x | x |} + 1} d x \dots . (i i) \end{array}

\begin{array}{l} 2 I = \int_{- 2}^{2} | x^{3} + x | d x \end{array}

\begin{array}{l} 2 I = 2 \int_{0}^{2} (x^{3} + x) d x \end{array}

\begin{array}{l} I = \int_{0}^{2} (x^{3} + x) d x \end{array}

Missing or unrecognized delimiter for \right

\begin{array}{l} = (\frac{16}{4} + \frac{4}{2}) - 0 \end{array}

= 4 + 2 = 6

11. If

\begin{array}{l} \frac{d y}{d x} + \frac{2^{x - y} (2^{y} - 1)}{2^{x} - 1} = 0, x, y > 0, y (1) = 1 \end{array}

, then y(2) is equal to :

(A) 2 + log₂3

(B) 2 + log₃2

(C) 2 – log₃2

(D) 2 – log₂3

Answer (D)

Sol.

\begin{array}{l} \frac{d y}{d x} + \frac{2^{x - y} (2^{y} - 1)}{2^{x} - 1} = 0 \end{array}

, x, y> 0, y(1) = 1

\begin{array}{l} \frac{d y}{d x} = - \frac{2^{x} (2^{y} - 1)}{2^{y} (2^{x} - 1)} \end{array}

\begin{array}{l} \int \frac{2^{y}}{2^{y} - 1} d y = - \int \frac{2^{x}}{2^{x} - 1} d x \end{array}

\begin{array}{l} = \frac{{log}_{e} (2^{y} - 1)}{{log}_{e} 2} = - \frac{{log}_{e} (2^{x} - 1)}{{log}_{e} 2} + \frac{{log}_{e} c}{{log}_{e} 2} \end{array}

= |(2^y – 1)(2^x – 1)| = c

∵ y(1) = 1

∴ c = 1

= |(2^y – 1)(2^x – 1)| = 1

For x = 2

|(2^y – 1)3| = 1

\begin{array}{l} 2^{y} - 1 = \frac{1}{3} \Rightarrow 2 y = \frac{4}{3} \end{array}

Taking log to base 2.

∴ y = 2 – log₂ 3

12. In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If (α, β) is the centroid of ΔABC, then 15(α + β) is equal to :

(A) 39

(B) 41

(C) 51

(D) 63

Answer (C)

Sol.

\begin{array}{l} \begin{array}{c} 2 x + y = 4 \\ 2 x + 6 y = 14 \end{array}} y = 2, x = 3 \end{array}

B(1, 2)

Let C(k, 4 – 2k)

Now AB² = AC²

5² + (–1)² = (6 – k)² + (–3 + 2k)²

⇒ 5k² – 24k + 19 = 0

(5k – 19)(k – 1) = 0

\begin{array}{l} \Rightarrow k = \frac{19}{5} \end{array}

\begin{array}{l} C (\frac{19}{5}, - \frac{18}{5}) \end{array}

Centroid (α, β)

\begin{array}{l} α = \frac{6 + 1 + \frac{19}{5}}{3} = \frac{18}{5} \end{array}

\(\begin{array}{l}\beta=\frac{1+2-\frac{18}{5}}{3}=-\frac{1}{5}\end{array} \)

Now 15(α + β)

\begin{array}{l} 15 (\frac{17}{5}) = 51 \end{array}

13. Let the eccentricity of an ellipse

\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b, \end{array}

be

\begin{array}{l} \frac{1}{4} \end{array}

. If this ellipse passes through the point

\begin{array}{l} (- 4 \sqrt{\frac{2}{5}}, 3) \end{array}

, then a² + b² is equal to :

(A) 29

(B) 31

(C) 32

(D) 34

Answer (B)

Sol.

\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{array}

\begin{array}{l} \Rightarrow \frac{{(- 4 \sqrt{\frac{2}{5}})}^{2}}{a^{2}} + \frac{32}{b^{2}} = 1 \end{array}

⇒

\begin{array}{l} \frac{32}{5 a^{2}} + \frac{9}{b^{2}} = 1 \end{array}

…(i)

a²(1 – e²) = b²

\begin{array}{l} a^{2} (1 - \frac{1}{16}) = b^{2} \end{array}

15a² = 16b² ⇒

\begin{array}{l} a^{2} = \frac{16 b^{2}}{15} \end{array}

From (i)

\begin{array}{l} \frac{6}{b^{2}} + \frac{9}{b^{2}} = 1 \Rightarrow b^{2} = 15 \end{array}

&a² = 16

a² + b² = 15 + 16 = 31

14. If two straight lines whose direction cosines are given by the relations l + m – n = 0, 3l² + m² + cnl = 0 are parallel, then the positive value of cis :

(A) 6

(B) 4

(C) 3

(D) 2

Answer (A)

Sol. l + m – n = 0 ⇒n = l + m

3l² + m² + cnl = 0

3l² + m² + cl(l + m) = 0

= (3 + c)l² + clm + m² = 0

\begin{array}{l} = (3 + c) {(\frac{I}{m})}^{2} + c (\frac{I}{m}) + 1 = 0 \end{array}

∵ Lines are parallel

D = 0

c² – 4(3 + c) = 0

c² – 4c – 12 = 0

(c – 4)(c + 3) = 0

c = 4 (as c> 0)

15. Let

\begin{array}{l} \vec{a} = \hat{i} + \hat{j} - \hat{k} \end{array}

and

\begin{array}{l} \vec{c} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k} \end{array}

. Then the number of vectors

\begin{array}{l} \vec{b} \end{array}

such that

\begin{array}{l} \vec{b} \times \vec{c} = \vec{a} \end{array}

and

\begin{array}{l} | \vec{b} | \in {1, 2, \dots, 10} \end{array}

is :

(A) 0

(B) 1

(C) 2

(D) 3

Answer (A)

Sol.

\begin{array}{l} \vec{a} = \hat{i} + \hat{j} - \hat{k} \end{array}

\begin{array}{l} \vec{c} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k} \end{array}

Now,

\begin{array}{l} \vec{b} \times \vec{c} = \vec{a} \end{array}

\begin{array}{l} \vec{c} \cdot (\vec{b} \times \vec{c}) = \vec{c} \cdot \vec{a} \end{array}

\begin{array}{l} \vec{c} \cdot \vec{a} = 0 \end{array}

\begin{array}{l} \Rightarrow (\hat{i} + \hat{j} - \hat{k}) (2 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 0 \end{array}

= 2 – 3 – 2 = 0

⇒ –3 = 0 (Not possible)

⇒ No possible value of

\begin{array}{l} \vec{b} \end{array}

is possible.

16. Five numbers, x₁, x₂, x₃, x₄, x₅ are randomly selected from the numbers 1, 2, 3,….., 18 and are arranged in the increasing order (x₁<x₂<x₃<x₄<x₅). The probability that x₂ = 7 and x₄ = 11 is:

(A)

\begin{array}{l} \frac{1}{136} \end{array}

(B)

\begin{array}{l} \frac{1}{72} \end{array}

(C)

\begin{array}{l} \frac{1}{68} \end{array}

(D)

\begin{array}{l} \frac{1}{34} \end{array}

Answer (C)

Sol. Total cases₌₁₈_C5

Favourable cases

\begin{array}{l} \begin{array}{c} ^{6} C_{1} & ^{3} C_{1} & ^{7} C_{1} \\ (Select x_{1}) & (Select x_{3}) & (Select x_{5}) \end{array} \end{array}

\begin{array}{l} P = \frac{6 \cdot 3 \cdot 7}{^{18} C_{5}} = \frac{1}{68} \end{array}

17. Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is:

(A)

\begin{array}{l} \frac{105}{16} \end{array}

(B)

\begin{array}{l} \frac{7}{16} \end{array}

(C)

\begin{array}{l} \frac{77}{36} \end{array}

(D)

\begin{array}{l} \frac{49}{16} \end{array}

Answer (C)

Sol. Given P(X = 3) = 5P(X = 4) and n = 7

⇒

\begin{array}{l} ^{7} C_{3} p^{3} q^{4} = 5 \cdot^{7} C_{4} p^{4} q^{3} \end{array}

⇒ q = 5p and also p + q = 1

⇒

\begin{array}{l} p = \frac{1}{6} and q = \frac{5}{6} \end{array}

_Mean

\begin{array}{l} = \frac{7}{6} \end{array}

_{and variance}

\begin{array}{l} = \frac{35}{36} \end{array}

Mean + Variance

\begin{array}{l} = \frac{7}{6} + \frac{35}{36} = \frac{77}{36} \end{array}

18. The value of

\begin{array}{l} \cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7}) \end{array}

is equal to:

(A) –1

(B)

\begin{array}{l} - \frac{1}{2} \end{array}

(C)

\begin{array}{l} - \frac{1}{3} \end{array}

(D)

\begin{array}{l} - \frac{1}{4} \end{array}

Answer (B)

Sol.

\begin{array}{l} \cos \frac{2 π}{7} + \cos \frac{4 π}{7} + \cos \frac{6 π}{7} = \frac{\sin 3 (\frac{π}{7})}{\sin \frac{π}{7}} \cos \frac{(\frac{2 π}{7} + \frac{6 π}{7})}{2} \end{array}

\begin{array}{l} = \frac{\sin (\frac{3 π}{7}) \cdot \cos (\frac{4 π}{7})}{\sin (\frac{π}{7})} \end{array}

\begin{array}{l} = \frac{2 \sin \frac{4 π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{π}{7}} \end{array}

\begin{array}{l} = \frac{\sin (\frac{8 π}{7})}{2 \sin \frac{π}{7}} = \frac{- \sin \frac{π}{7}}{2 \sin \frac{π}{7}} = \frac{- 1}{2} \end{array}

19.

\begin{array}{l} \sin^{- 1} (\sin \frac{2 π}{3}) + \cos^{- 1} (\cos \frac{7 π}{6}) + \tan^{- 1} (\tan \frac{3 π}{4}) \end{array}

is equal to:

(A)

\begin{array}{l} \frac{11 π}{12} \end{array}

(B)

\begin{array}{l} \frac{17 π}{12} \end{array}

(C)

\begin{array}{l} \frac{31 π}{12} \end{array}

(D)

\begin{array}{l} - \frac{3 π}{4} \end{array}

Answer (A)

Sol.

\begin{array}{l} \sin^{- 1} (\frac{\sqrt{3}}{2}) + \cos^{- 1} (\frac{- \sqrt{3}}{2}) + \tan^{- 1} (- 1) \end{array}

\begin{array}{l} = \frac{π}{3} + \frac{5 π}{6} - \frac{π}{4} \end{array}

\begin{array}{l} = \frac{4 π + 10 π - 3 π}{12} = \frac{11 π}{12} \end{array}

20. The boolean expression (~(p ∧q)) ∨q is equivalent to:

(A) q→ (p ∧q)

(B) p→q

(C) p→ (p→q)

(D) p→ (p∨q)

Answer (D)

Sol. Making truth table

p	q	p ∧ q	~p ∧ q	(~(p ∧ q)) ∨ q	p ∨ q	p → q	p → (p ∨ q)
T	T	T	F	T	T	T	T
T	F	F	T	T	T	F	T
F	T	F	T	T	T	T	T
F	F	F	T	T	F	T	T
				Tautology			Tautology

\begin{array}{l} ∴ (\sim (p \land q)) \lor q \equiv p \to (p \lor q) \end{array}

SECTION – B

Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

1. Let ƒ :R→R be a function defined by

\begin{array}{l} f (x) = \frac{2 e^{2 x}}{e^{2 x} + e^{x}} \end{array}

Then

\begin{array}{l} f (\frac{1}{100}) + f (\frac{2}{100}) + f (\frac{3}{100}) + \dots + f (\frac{99}{100}) \end{array}

is equal to ________.

Answer (99)

Sol.

\begin{array}{l} f (x) = \frac{2 e^{2 x}}{e^{2 x} + e^{x}} and f (1 - x) = \frac{2 e^{2 - 2 x}}{e^{2 - 2 x} + e^{1 - x}} \end{array}

∴

\begin{array}{l} \frac{f (x) + f (1 - x)}{2} = 1 \end{array}

i.e. f(x) + f(1 – x) = 2

∴

\begin{array}{l} f (\frac{1}{100}) + f (\frac{2}{100}) + \dots + f (\frac{99}{100}) \end{array}

\begin{array}{l} = \sum_{x = 1}^{49} f (\frac{x}{100}) + f (1 - \frac{x}{100}) + f (\frac{1}{2}) \end{array}

= 49 × 2 + 1 = 99

2. If the sum of all the roots of the equation

\(\begin{array}{l}e^{2x} – 11e^x – 45e^{–x} +\frac{81}{2}=0\end{array} \)

is log_ep, then p is equal to _____.

Answer (45)

Sol. Let e^x = t then equation reduces to

\begin{array}{l} t^{2} - 11 t - \frac{45}{t} + \frac{81}{2} = 0 \end{array}

⇒ 2t³ – 22t² + 81t – 45 = 0 …(i)

if roots of

\begin{array}{l} e^{2 x} - 11 e^{x} - 45 e^{- x} + \frac{81}{2} = 0 \end{array}

are α, β, γ then roots of (i) will be

\begin{array}{l} e^{α_{1}} e^{α_{2}} e^{α_{3}} \end{array}

using product of roots

\begin{array}{l} e^{α_{1} + α_{2} + α_{3}} = 45 \end{array}

⇒ α₁ + α₂ + α₃ = ln 45 ⇒ p = 45

3. The positive value of the determinant of the matrix A, whose

\begin{array}{l} A d j (A d j (A)) = (\begin{array}{c} 14 & 28 & - 14 \\ - 14 & 14 & 28 \\ 28 & - 14 & 14 \end{array}) \end{array}

, is __________.

Answer (14)

Sol.

\begin{array}{l} | a d j (a d j (A)) | = {| A |}^{2^{2}} = {| A |}^{4} \end{array}

∴

\begin{array}{l} {| A |}^{4} = | \begin{array}{c} 14 & 28 & - 14 \\ - 14 & 14 & 28 \\ 28 & - 14 & 14 \end{array} | \end{array}

\begin{array}{l} = {(14)}^{3} | \begin{array}{c} 1 & 2 & - 1 \\ - 1 & 1 & 2 \\ 2 & - 1 & 1 \end{array} | \end{array}

\begin{array}{l} = {(14)}^{3} (3 - 2 (- 5) - 1 (- 1)) \end{array}

\begin{array}{l} {| A |}^{4} = {(14)}^{4} \Rightarrow | A | = 14 \end{array}

4. The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is _________.

Answer (56)

Sol. First we arrange 5 red cubes in a row and assume x₁, x₂, x₃, x₄, x₅ and x₆ number of blue cubes between them

Here, x₁+ x₂+ x₃+ x₄+ x₅ + x₆= 11

and x₂, x₃, x₄, x₅≥ 2

So x₁+ x₂+ x₃+ x₄+ x₅ + x₆= 3

No. of solutions = ⁸C₅ = 56

5. If the coefficient of x¹⁰ in the binomial expansion of

\begin{array}{l} {(\frac{\sqrt{x}}{5^{\frac{1}{4}}} + \frac{\sqrt{5}}{x^{\frac{1}{3}}})}^{60} \end{array}

is 5^k^.l, where l, k∈N and l is co-prime to 5, then k is equal to ___________.

Answer (5)

Sol.

\begin{array}{l} T_{r + 1} = 60_{C_{r}} {(x^{\frac{1}{2}})}^{60 - r} {(x^{- \frac{1}{3}})}^{r} {(5^{\frac{- 1}{4}})}^{60 - r} {(5^{\frac{1}{2}})}^{r} \end{array}

for

\begin{array}{l} x^{10} \frac{60 - r}{2} - \frac{r}{3} = 10 \end{array}

⇒ 180 – 3r – 2r = 60

⇒ r = 24

∴ Coeff. of

\(\begin{array}{l}x^{10}=\frac{^{60}C_{24}}{5^9} 5^{12}=5^k~~I\end{array} \)

as l and 5 are coprime

k = 3 + exponent of 5 in ⁶⁰C₂₄

\begin{array}{l} = 3 + ([\frac{60}{5}] + [\frac{60}{5^{2}}] - [\frac{24}{5}] - [\frac{24}{5^{2}}] - [\frac{36}{5}] - [\frac{36}{5^{2}}] \end{array}

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

6. Let

\begin{array}{l} A_{1} = {(x, y) : | x | \leq y^{2}, | x | + 2 y \leq 8} \end{array}

and

\begin{array}{l} A_{2} = {(x, y) : | x | + | y | \leq k} \end{array}

. If 27(Area A₁) = 5(Area A₂), then k is equal to :

Answer (6*)

Sol.

Required area (above x-axis)

\begin{array}{l} A_{1} = 2 \int_{0}^{4} (\frac{8 - x}{2} - \sqrt{x}) d x \end{array}

\begin{array}{l} = 2 (16 - \frac{16}{4} - \frac{8}{3 / 2}) = \frac{40}{3} \end{array}

and

\begin{array}{l} A_{2} = 4 (\frac{1}{2} \cdot k^{2}) = 2 k^{2} \end{array}

JEE Main 2022 June 27 Shift 1 Maths NQA 6(ii)

∴

\begin{array}{l} 27 \cdot \frac{40}{3} = 5 \cdot (2 k^{2}) \end{array}

⇒ k = 6

* A₁ is

JEE Main 2022 June 27 Shift 1 Maths NQA 6(iii)

Which tends to infinity if not mentioned above x-axis

7. If the sum of the first ten terms of the series

\begin{array}{l} \frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \dots \end{array}

is

\begin{array}{l} \frac{m}{n} \end{array}

, where m and n are co-prime numbers, then m + n is equal to __________.

Answer (276)

Sol.

\begin{array}{l} T_{r} = \frac{r}{{(2 r^{2})}^{2} + 1} \end{array}

\begin{array}{l} = \frac{r}{{(2 r^{2} + 1)}^{2} - {(2 r)}^{2}} \end{array}

\begin{array}{l} = \frac{1}{4} \frac{4 r}{(2 r^{2} + 2 r + 1) (2 r^{2} - 2 r + 1)} \end{array}

\begin{array}{l} S_{10} = \frac{1}{4} \sum_{r = 1}^{10} (\frac{1}{(2 r^{2} - 2 r + 1)} - \frac{1}{(2 r^{2} + 2 r + 1)}) \end{array}

\begin{array}{l} = \frac{1}{4} [1 - \frac{1}{5} + \frac{1}{5} - \frac{1}{13} + \dots \dots + \frac{1}{181} - \frac{1}{221}] \end{array}

⇒

\begin{array}{l} S_{10} = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221} = \frac{m}{n} \end{array}

∴ m + n = 276

8. A rectangle R with end points of one of its sides as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is

2x – y + 4 = 0, then the area of R is ________.

Answer (16)

Sol.

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side

∴

\begin{array}{l} a = \sqrt{{(3 - 1)}^{2} + {(6 - 2)}^{2}} = \sqrt{20} \end{array}

and

\begin{array}{l} \frac{b}{2} = \frac{4}{\sqrt{5}} \Rightarrow b = \frac{8}{\sqrt{5}} \end{array}

Area

\begin{array}{l} a b = 2 \sqrt{5} \cdot \frac{8}{\sqrt{5}} = 16 \end{array}

.

9. A circle of radius 2 unit passes through the vertex and the focus of the parabola y² = 2x and touches the parabola

\begin{array}{l} y = {(x - \frac{1}{4})}^{2} + α \end{array}

where α> 0. Then (4α – 8)² is equal to ___________.

Answer (63)

Sol.

Let the equation of circle be

\begin{array}{l} x (x - \frac{1}{2}) + y^{2} + λ y = 0 \end{array}

⇒

\begin{array}{l} x^{2} + y^{2} - \frac{1}{2} x + λ y = 0 \end{array}

Radius

\begin{array}{l} = \sqrt{\frac{1}{16} + \frac{λ^{2}}{4}} = 2 \end{array}

⇒

\begin{array}{l} λ^{2} = \frac{63}{4} \end{array}

⇒

\begin{array}{l} {(x - \frac{1}{4})}^{2} + {(y + \frac{λ}{2})}^{2} = 4 \end{array}

∵ This circle and parabola

\begin{array}{l} y - α = {(x - \frac{1}{4})}^{2} \end{array}

touch each other, so

\begin{array}{l} α = - \frac{λ}{2} + 2 \end{array}

⇒

\begin{array}{l} α - 2 = - \frac{λ}{2} \end{array}

⇒

\begin{array}{l} {(α - 2)}^{2} = \frac{λ^{2}}{4} = \frac{63}{16} \end{array}

⇒

\begin{array}{l} {(4 α - 8)}^{2} = 63 \end{array}

10. Let the mirror image of the point (a, b, c) with respect to the plane 3x – 4y + 12z + 19 = 0 be

(a – 6, β, γ). If a + b + c = 5, then 7β – 9γ is equal to __________.

Answer (137)

Sol.

\begin{array}{l} \frac{x - a}{3} = \frac{y - b}{- 4} = \frac{z - c}{12} = \frac{- 2 (3 a - 4 b + 12 c + 19)}{3^{2} + {(- 4)}^{2} + 12^{2}} \end{array}

\begin{array}{l} \frac{x - a}{3} = \frac{y - b}{- 4} = \frac{z - c}{12} = \frac{- 6 a + 8 b - 24 c - 38}{169} \end{array}

\begin{array}{l} (x, y, z) \equiv (a - 6, β, γ) \end{array}

\begin{array}{l} \frac{(a - b) - a}{3} = \frac{β - b}{- 4} = \frac{γ - c}{12} = \frac{- 6 a + 8 b - 24 c - 38}{169} \end{array}

\begin{array}{l} \frac{β - b}{- 4} = - 2 \end{array}

⇒ β = 8 + b

\begin{array}{l} \frac{γ - c}{12} = - 2 \end{array}

⇒ γ = -24 + c

\begin{array}{l} \frac{- 6 a + 8 b - 24 c - 38}{169} = - 2 \end{array}

⇒ 3a – 4b + 12c = 150….(1)

a + b + c = 5

⇒ 3a + 3b + 3c = 15….(2)

Applying (1) – (2), we get;

-7b + 9c = 135

7b – 9c = -135

7β – 9γ = 7(8 + b) – 9 (–24 + c)

= 56 + 216 + 7b – 9c.

= 56 + 216 – 135 = 137.

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JEE Main 2022 June 27th Shift 1 Question Paper & Solutions

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Frequently Asked Questions – FAQs

Q1

How was the JEE Main 2022 June 27 Shift 1 Maths question paper?

The JEE Main 2022 June 27 Shift 1 Maths question paper was easy to moderate level difficulty. There were 8 easy questions, 11 medium questions and 2 difficult questions in the JEE Main 2022 June 27 Shift 1 Maths question paper.

Q2

Is there a negative marking in section B of the JEE Main 2022 June 27 Maths Shift 1 question paper?

No. There is no negative marking in section B of the JEE Main 2022 June 27 Maths Shift 1 question paper.

Q3

How many questions were asked from Class 11 in the JEE Main 2022 Maths June 27 Shift 1 question paper?

There were 10 questions from Class 11 in the JEE Main 2022 Maths June 27 Shift 1 question paper as per the memory-based question paper.

Q4

What is the overall difficulty level of the JEE Main 2022 Maths June 27 Shift 1 question paper?

The overall difficulty level of JEE Main 2022 June 27 Shift 1 Maths question paper is 1.71 out of 3

Q5

Were there any questions from Quadratic equations in the JEE Main 2022 June 27 Maths Shift 1 question paper?

Yes. Questions were there from Quadratic equations in the JEE Main 2022 June 27 Maths Shift 1 question paper, and those were easy.

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